Identifying Probability Distributions. In Exercises 7–14, dete...

Question

Identifying Probability Distributions. In Exercises 7–14, determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied.
Fear of Heights The table lists results from a survey of 285 subjects who were asked, “Are you afraid of heights in tall buildings?” The results are from USA Today.

Fear of Heights The table lists results from a survey of 285 subjects who were asked, “Are you afraid of heights in tall buildings?” The results are from USA Today.

Table showing survey results:

Accepted Answer

Hello everyone. Let's take a look at this question together. A health magazine conducted a poll among 800 readers about their weekly exercise habits. The question was, do you exercise more than 3 times a week? The responses were collected as shown below. Determine if the table represents a probability distribution, and if so, find its mean and standard deviation. And here we have our table where the response is represented by the variable X, and we have the number of readers, and the probability that the random variable X has a value that is equal to our response variable X. So in order to solve this question, we must first recall what we have learned about a probability distribution to determine if this table represents a probability distribution, and if so, find the mean and the standard deviation. And so the first step in solving this problem is determining if the table represents a probability distribution, which we can do so by checking the two key conditions, starting with the first key condition, which is that each probability is between 0 and 1, which looking at our table, we can determine that the answer to the first key condition is yes, and the second key condition is that The sum of the probabilities is equal to 1, which from the table we see our probabilities are 0.7 and 0.3, and the sum of 0.7 plus 0.3 is in fact equal to 1. So based on our data for the second key condition, we know that it is true. Therefore, this is a valid probability distribution, so we now need to find the mean and the standard deviation. And so the next step is to define the random variable, such as the random variable X is equal to 1 if the person exercises more than 3 times a week, and this will correspond to our yes or our response variable of our lowercase x. And then our random. variable X will equal 0 if the person does not exercise more than 3 times a week, which corresponds to our no or our response variable X. And now we can find the mean, which the mean can be found using the formula mu is equal to the sum of X multiplied by P X. And using the values in our table, as well as our definition of the random variable, we can substitute the values into our mean formula, which plugging in our values into the formula, we have one multiplied by 0.7 plus 0 multiplied by 0.3. So the calculation for this problem results in our mean of 0.7. Therefore, we have a mean of mu equals 0.7. And now we can find our standard deviation, which can be calculated using the formula for standard deviation where sigma squared is equal to the sum of X minus mu 2 multiplied by p of X, so substituting the values into our formula for the standard. we have sigma squared is equal to in parentheses 1 minus 0.7 squared multiplied by 0.7 plus in parentheses 0 minus 0.7 squared multiplied by 0.3. And when simplifying this data, our calculation for sigma squared is equal to 0.21. However, we are looking for the standard deviation and sigma squared is equal to the standard variance. So to find our standard deviation, we need the square root of sigma squared, so we plug in our value of 0.21 into the formula to find the standard deviation. Which the square root of 0.21 is approximately 0.4583. Therefore, we have a mean of 0.7 and a standard deviation of approximately 0.4583. So our final answer is that this table does represent a probability distribution, and our mean is 0.7, and our standard deviation is 0.46. So the correct answer choice is answer choice D. Yes, it is a valid probability distribution with a mean of 0.7 and a standard deviation of 0.46. I hope you found this video to be helpful. Thank you and goodbye.

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